Combustion analysis of a 13.42-g sample of equilin (which contains only carbon, hydrogen, and oxygen) produces 39.61 g CO 2 and 9.01 g H 2 O. The molar mass of equilin is 268.34 g/mol. Find its molecular formula.

Hello, everyone. Today we have the following problem. Combustion analysis of a 9.26 g sample of Trend ion produces 27.38 g of carbon dioxide and 6.22 g of water. The molar mass of Trend ion is 2 68.34 g from find its molecular formula. So for a combustion reaction, our general reaction would look like the following. So we would have the hydrocarbon reacting with our oxygen gas to form X amount of moles of our carbon dioxide and y amount of moles of our water. So we need to calculate the mass of our carbon and hydrogen in the compound first. So our mouse of carbon and grams of course will be equal to, we have to start with our initial value of 27.34 g of carbon dioxide to get from carbon dioxide to carbon. We first need to multiply by the molar mass of carbon dioxide which is two oxygens and one carbon. And if we add up their individual masses, we get 44.01 g is equal to one more. And then we have to multiply by the multiple ratio. And in our equation, we said that one mole of CO2 was equal to one mole of carbon produced. And then lastly, we multiply our one mole of carbon by its molar mass which is 12.01 g. When our units canceled out, we arrive at 7.46 g. No, we calculate our mass of hydrogen. We started with 6.22 g. We multiply bytes mola mass of water, which is one mole is equal to two hydrogens and one oxygen for 18.02 g. It will be multiplied by the multiple ratio. So one mole of water was produced for the use of two moles of our hydrogen. So we see that in our water, we have the two subscript for the hydrogen. That means we have two moles of it. And then lastly, we multiply by the molar mass, one mole of hydrogen is equal to 1.01 g of hydrogen. Our units canceled out and we can solve for a mass of 0.97 g of hydrogen. So we have 9.26 g of the sample compound. And we need to essentially subtract that to find the mass of oxygen. So the mass of oxygen would be equal to our 9.26 g. We subtract that from our mass of carbon 7.46 and as well as our mass of our hydrogen, which is 0.697 g for a mass of 1.103 g of oxygen. Next, we need to calculate the moles of each element and determine the lowest hole number ratio. So, moles of carbon will be equal to the mass that we have, which is 7.46 g multiplying by its molar mass for 0.621 we do the same for our hydrogen. We at 0.697 g multiplying by its smaller mass for a total of 0.690. And then lastly, we have our moles of oxygen. It should be 1.103 g multiplied by its molar mass of 16 g, 40 12.069. Then we need to divide the lowest divide the number of moles of our carbon hydrogen and oxygen by the smallest value which we have for moles, which would be the 0.069. So we do 0.069 for all of them. And we arrive at nine moles for carbon 10 moles of hydrogen and one mole for our oxygen. And so the empirical formula, we can say that the empirical formula is C nine H 10 0. To determine the mole molecular formula from that, we divide the given molar mass by the mass of the empirical formula. So we have nine moles of carbon. We multiply that by the molar mass, we do the same thing for our hydrogen and or oxygen. We get 108.9 g from all for carbon 10.10 g per mo per hour hydrogen. And then we have 16 g per mole for our oxygen. Now, the molar mass divided by the empirical mass gives us the ratio in which we can multiply our values for. So we have our molar mass divided by our empirical mass that is 2 64.34 g per mole for our molar mass. And then we have our 1 34.19 g small for our empirical mass giving us a ratio of about two. So we essentially need to multiply these values by two such that our molecular formula can be C 18 H 2002. And if you look at our anti choice, because we see Anto D best reflects this overall, I hope it helped. And until next time.

Ch.3 - Molecules and Compounds

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Tro 6th Edition

Ch.3 - Molecules and Compounds

Problem 129

Chapter 3, Problem 129

Combustion analysis of a 13.42-g sample of equilin (which contains only carbon, hydrogen, and oxygen) produces 39.61 g CO₂ and 9.01 g H₂O. The molar mass of equilin is 268.34 g/mol. Find its molecular formula.

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Calculate the moles of CO2 produced using its molar mass: \( \text{moles of CO}_2 = \frac{39.61 \text{ g}}{44.01 \text{ g/mol}} \).

Determine the moles of carbon in the sample from the moles of CO2, knowing that each mole of CO2 contains one mole of carbon.

Calculate the moles of H2O produced using its molar mass: \( \text{moles of H}_2\text{O} = \frac{9.01 \text{ g}}{18.02 \text{ g/mol}} \).

Determine the moles of hydrogen in the sample from the moles of H2O, knowing that each mole of H2O contains two moles of hydrogen.

Calculate the mass of carbon and hydrogen in the sample, subtract from the total mass to find the mass of oxygen, and then determine the moles of oxygen.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Combustion Analysis

Combustion analysis is a technique used to determine the elemental composition of organic compounds. In this process, a sample is burned in excess oxygen, producing carbon dioxide and water. The masses of these products are measured, allowing for the calculation of the amounts of carbon, hydrogen, and oxygen in the original sample.

Empirical Formula

The empirical formula represents the simplest whole-number ratio of the elements in a compound. To find it, the moles of each element are calculated from the combustion products, and then the ratios are simplified. This formula is crucial for determining the molecular formula when combined with the molar mass of the compound.

Molecular Formula

The molecular formula indicates the actual number of atoms of each element in a molecule of a compound. It can be derived from the empirical formula by multiplying it by a whole number, which is determined by dividing the compound's molar mass by the molar mass of the empirical formula. This provides insight into the compound's structure and properties.